from functools import reduce def str2float(s): DIGITS = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9, '.': '.'} def str2int(s): return DIGITS[s] def f1(s): return s.index('.') #返回 L中'' .'' 的位置 def fn1(x, y): return x*10+y ans = reduce(fn1,map(str2int,(s[:f1(s)]+s[f1(s)+1:]))) #s[:f1(s)]+s[f1(s)+1:] 对原字符串进行切片重组 字符串连接可以用+ return ans*pow(0.1,(len(s)-1-f1(s)))
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陳大發的小红豆